/**
 * @Author liangzai
 * @Description:
 */
public class test {
    public ListNode head;

    public boolean contains(int key) {
        ListNode cur = head;
        while (cur != null) {
            if(cur.val == key) {
                return true;
            }
            cur = cur.next;
        }
        return false;
    }



    public ListNode partition(int x) {
        // write code here
        ListNode bs = null;
        ListNode be = null;
        ListNode as = null;
        ListNode ae = null;
        ListNode cur = head;
        while (cur != null) {
            if(cur.val < x) {
                //第一次插入
                if(bs == null) {
                    bs = be = cur;
                }else {
                    be.next = cur;
                    be = be.next;
                }
            }else {
                if(as == null) {
                    as = ae = cur;
                }else {
                    ae.next = cur;
                    ae = ae.next;
                }
            }
            cur = cur.next;
        }
        if(bs == null) {
            return as;
        }
        be.next = as;
        //不管第二部分是否最后一个节点为空 都手动置为空
        if(as != null) {
            ae.next = null;
        }
        return bs;
    }

    //10:13 我来写
    public boolean chkPalindrome() {
        // write code here
        if(head == null) return true;
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        //slow 指向的位置 就是中间节点
        //2.进行翻转
        ListNode cur = slow.next;
        while (cur != null) {
            ListNode curN = cur.next;
            cur.next = slow;
            slow = cur;
            cur = curN;
        }
        //3.判断回文
        while (head != slow) {
            if(head.val != slow.val) {
                return false;
            }
            if(head.next == slow) {
                return true;
            }
            head = head.next;
            slow = slow.next;
        }
        return true;
}
